To measured a magnetic field between the magnetic poles of a loud speaker, a small coil having 30 turns and…
To measured a magnetic field between the magnetic poles of a loud speaker, a small coil having 30 turns and \(2.5 \mathrm{~cm}^2\) area is placed perpendicular to the field and removed immediately. If the total charge flown through the coil is \(75 \times 10^{-3} \mathrm{C}\) and the total resistance of wire and galvanometer is \(0.3 \Omega\), then the magnitude of the magnetic field is
\(0.03 \mathrm{~T}\)
\(0.3 \mathrm{~T}\)
\(3 \mathrm{~T}\)
\(3 \times 10^2 \mathrm{~T}\)
Solution
Given,
number of turns in the coil, \(N=30\) turns
area of the coil, \(A=2.5 \mathrm{~cm}^2=2.5 \times 10^{-4} \mathrm{~m}^2\)
total charge flowing through the coil,
\(Q_{\text {Net }}=7.5 \times 10^{-3} \mathrm{C}\)
and total resistance of wire and galvanometer,
\(R=0.3 \Omega\)
We know that,
\(\text { net charge, } Q_{n e t}=\frac{\text { flux in the coil }}{\text { net Resistance }}\)
\(\therefore \quad Q_{\text {net }}=\frac{\phi}{R}\) ...(i)
We know that, flux \(\phi=\) magnetic field \(\times\) area
\(\times\) number of turns in the coil
or \(\phi=B N A\) ...(ii)
Now, from Eqs. (i) and (ii), we get
\(\therefore \quad Q_{\text {Net }}=\frac{B N A}{R}\)
Putting the given values, we get
\(\begin{aligned}
7.5 \times 10^{-3} & =\frac{B \times 30 \times\left(2.5 \times 10^{-4}\right)}{0.3} \\
7.5 \times 10^{-3} & =\frac{B \times 7.5 \times 10^{-3}}{0.3} \\
B & =0.3 \mathrm{~T}
\end{aligned}\)
Hence, the magnitude of the magnetic field, \(B=0.3 \mathrm{~T}\).